AP EAMCET · Maths · Straight Lines
\(P\) is a variable point such that the distance of \(P\) from \(A(4,0)\) is twice the distance of \(P\) from \(B(-4,0)\). If the line \(3 y-3 x-20=0\) intersects the locus of P at the points C and D , then the distance between C and D is
- A 8
- B \(\frac{8 \sqrt{2}}{3}\)
- C \(\frac{32}{3}\)
- D \(\frac{8}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{32}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { } \sqrt{(x-4)^2+y^2}=2 \sqrt{(x+4)^2+y^2} \\ & \Rightarrow 3 x^2+3 y^2+40 x+48=0 \\ & \text { Centre } \equiv\left(\frac{-20}{3}, 0\right) ; \text { Radius }=\sqrt{\left(\frac{20}{3}\right)^2-16}=\frac{16}{3} \end{aligned}\) Given line is…
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